Question

For the reaction; $N_2+3H_2\to 2N{H}_3$, the rate $\frac{d\left[N{H}_3\right]}{dt}=2\times {10}^{-4}M{s}^{-1}$. Therefore, the rate $-\frac{d\left[N_2\right]}{dt}$ is given as:

• A. ${10}^{-4}M{s}^{-1}$
• B. ${10}^{4}M{s}^{-1}$
• C. ${10}^{-2}M{s}^{-1}$
• D. ${10}^{-4}s{M}^{-1}$

Answer 1

Answer: (A)

${10}^{-4}M{s}^{-1}$

$N_2+3H_2\to 2N{H}_3$
Rate =
$\frac{-d\left[N_2\right]}{dt}=\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}=\frac{2\times {10}^{-4}}{2}={10}^{-4}M{s}^{-1}$